Question

If $ \mu_0 $ and $ \epsilon_0 $ are the permeability and permittivity of free space, respectively, then the dimension of $ \left( \frac{1}{\mu_0 \epsilon_0} \right) $ is :

• A. \( L T^2 \)
• B. \( L^2 T^{-2} \)
• C. \( T^2 / L \)
• D. \( T^2 / L^2 \)

Hint:

The dimensions of permeability and permittivity help determine the speed of light in a vacuum, and their relationship is key in electromagnetic theory.

Answer 1

The Correct Option is B

Determine the dimensions of \( \left(\dfrac{1}{\mu_0 \epsilon_0}\right) \), where \( \mu_0 \) denotes the permeability and \( \epsilon_0 \) denotes the permittivity of free space.

Concept Used:

The relationship between the speed of light \( c \), permeability \( \mu_0 \), and permittivity \( \epsilon_0 \) is given by:

\[c = \frac{1}{\sqrt{\mu_0 \epsilon_0}}\]

Rearranging this, we get:

\[\frac{1}{\mu_0 \epsilon_0} = c^2\]

The dimensional formula for speed \( c \) is \( [L\,T^{-1}] \). Therefore, the dimensions of \( c^2 \) are \( [L^2\,T^{-2}] \).

Step-by-Step Solution:

Step 1: State the dimensional formulas for \( \mu_0 \) and \( \epsilon_0 \):

\[[\mu_0] = [M^1 L^1 T^{-2} A^{-2}]\]\[[\epsilon_0] = [M^{-1} L^{-3} T^{4} A^{2}]\]

Step 2: Calculate the product \( \mu_0 \epsilon_0 \):

\[[\mu_0 \epsilon_0] = [M^{1-1} L^{1-3} T^{-2+4} A^{-2+2}] = [L^{-2} T^{2}]\]

Step 3: Compute the reciprocal:

\[\left[\frac{1}{\mu_0 \epsilon_0}\right] = [L^2 T^{-2}]\]

Final Computation & Result:

The dimensions of \( \left(\dfrac{1}{\mu_0 \epsilon_0}\right) \) are:

\[\boxed{[L^2 T^{-2}]}\]

This result corresponds to the square of the velocity, \( c^2 \).